Chap3StudentSolutions

# For n 15 y n 2m 2m 7 sin 8 m 7 sin 8 0 m

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Unformatted text preview: r &lt; 1 from Appendix A, 1− r y[ n ] = (0.95) (b) (0.95) n − m u[n − m] −q −n Using 2πm 32 m =−∞ (0.95) n − m m =−∞ j 2πn x[ n ] = sin 32 2π −j 0.95e 32 n 1 − 0.95e −j −n 2π 32 = 2π n 32 1 − 0.95e −j 2π 32 = 5.0632e 2π j n −1.218 32 h[ n ] = (0.95) u[ n ] n , From part (a), the response to x[ n ] = e e j j 2πn 32 is y[ n ] = 5.0632e 2πn e sin = 32 j 2πn 32 −e j2 −j 2π j n −1.218 32 2πn 32 , Solutions 3-13 . Since M. J. Roberts - 8/16/04 2πn by applying linearity and superposition, the response to x[ n ] = sin is 32 y[ n ] = 5.0632e 2π j n −1.218 32 − 5.0632e j2 2π − j n −1.218 32 16. Given the excitations, x[ n ] , and the impulse responses, h[ n ], use MATLAB to plot the system responses, y[ n ] . 2πn , h[ n ] = sin (a) x[ n ] = u[ n ] − u[ n − 8] (u[ n ] − u[ n − 8]) 8 y[ n ] = h[ n ] ∗ x[ n ] = y[ n ] = ∞ 2πm (u[ m] − u[ m − 8])(u[ n − m] − u[ n − m − 8]) 8 ∞ ∑ sin m =−∞ 2πm u[ m] u[ n − m] − u[ m] u[ n − m − 8] 8 − u[ m − 8] u[ n − m] + u[ m − 8] u[ n − m − 8] ∑ sin m =−∞ 2πm 2πm 2πm 2πm y[ n ] = ∑ sin + ∑ sin − ∑ sin − ∑ sin 8 m =0 8 m =8 8 m =8 8 m =0 n −8 n n −8 n For n &lt; 0, all the summations are zero because the factors, u [ n − m ] and u [ n − m − 8 ] are zero in the summation range, 0 &lt; m &lt; n , and y[ n ] = 0. For n &gt; 15, y[ n ] = 2πm 2πm ∑− 7 sin 8 − m∑− 7 sin 8 = 0 . m =n =n n n So the response is only non-zero for 0 ≤ m &lt; 16 (and can be zero at some points within that range). (b) 2πn x[ n ] = sin (u[ n ] − u[ n − 8]) 8 2πn h[ n ] = − sin (u[ n ] − u[ n − 8]) 8 , 17. Which of these systems are BIBO stable? (a) x[n] y[n] -0.9 D The system equation is y[ n ] = x[ n ] − 0.9 y[ n − 1] Solutions 3-14 M. J. Roberts - 8/16/04 The eigenvalue is α = −0.9 . Its magnitude is less than one, therefore the system is stable. (b) x[n] y[n] D 1.1 (c) x[n] y[n] 1 2 D 1 2 D (d) x[n] y[n] 1.5 D 0.4 D 18. Find and plot the unit-sequence responses of these systems. (a) x[n] y[n] D 0.7 D -0.5 h[ n ] = h1[ n ] ∗ h 2 [ n ] h1[ n ] = (0.7) u[ n ] n h[ n ] = ∞ and h 2 [ n ] = (−0.5) u[ n ] ∑ (0.7) u[m](−0.5) m =−∞ m n n −m u[ n − m] Simplify this expression as much as possible by letting the unit sequency functions modify the summation limits and then apply the formular for the summation of a geometric series, Solutions 3-15 M. J. Roberts - 8/16/04 1 ∑ r = 1 − r N n=0 1− r , N −1 r =1 n to get , otherwise r 1 − ( −1.4 ) h [ n ] = ( −0.5 ) 2.4 n +1 n u[n] Then convolve the impulse response with the unit sequence to get the overall response and use some of the same techniques to find a simple closed-form expression for the response. { ( y[ n ] = 0.4167 0.6667 1 − (−0.5) (b) x[n] n +1 ) + 4.6667(1 − (0.7) )} u[n] n +1 D -0.8 y[n] D D 0.6 h[ n ] = h1[ n ] + h 2 [ n ] ( ) ( ) n n n h[ n ] = (−0.8) + 0.6455 0.6 − 0.6455 − 0.6 u[ n ] Then convolve...
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## This note was uploaded on 06/19/2013 for the course ENSC 380 taught by Professor Atousa during the Spring '09 term at Simon Fraser.

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